Proof. Let $R$ be a principal ideal domain. Let $g = a/b$ be an element of the fraction field of $R$ integral over $R$. Because $R$ is a principal ideal domain we may divide out a common factor of $a$ and $b$ and assume $(a, b) = R$. In this case, any equation $(a/b)^ n + r_{n-1} (a/b)^{n-1} + \ldots + r_0 = 0$ with $r_ i \in R$ would imply $a^ n \in (b)$. This contradicts $(a, b) = R$ unless $b$ is a unit in $R$. $\square$

Comment #5508 by Manuel Hoff on

The same proof shows that in fact any factorial ring is normal. One just has to replace $(a, b) = R$ with the (weaker) condition that $a$ and $b$ are coprime in the sense that their greatest common divisor is $1$. This is enough to make the conclusion.

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• 3 comment(s) on Section 10.37: Normal rings

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